Question

Write a C++ program to calculate cube of first 15 even numbers.

Answer 1

The efficient approach is discussed below

The sum of cubes of first n natural numbers is given by = (n*(n+1) / 2)^2

Sum of cubes of first n natural numbers can be written as...

= 2^3 + 4^3 + .... + (2n)^3

Now take out common term i.e 2^3

= 2^3 * (1^3 + 2^3 + .... + n^3)

= 2^3* (n*(n+1) / 2)^2

= 8 * ((n^2)(n+1)^2)/4

= 2 * n^2(n+1)^2

#include <iostream>

using namespace std;

int calculate(int n)

{

int sum = 2 * n * n * (n + 1) * (n + 1);

return sum;

}

int main()

{

int num = 15;

cout<<"Number is = "<<num<<endl;

cout << "Sum of cubes of first "<<num<<" even number is ="<<calculate(num);

return 0;

}

Answer 2

Answer:

Answer:

#include <iostream>

using namespace std;

int calculate(int n)

{

int sum = 2 * n * n * (n + 1) * (n + 1);

return sum;

}

int main()

{

int num = 15;

cout<<"Number is = "<<num<<endl;

cout << "Sum of cubes of first "<<num<<" even number is ="<<calculate(num);

return 0;

}

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